[obm-l] Uma soluçao bonitinha da IMO da India(de novo)

[peterdirichlet@xxxxxxxxxxxxxx]

Prova do lema:veja que os triangulos DEF e DMF sao semelhantes,logo tem
o mesmo circunraio.Mas XM e o diametro do triangulo DMF.Logo XM=2*RA.E reformule
a desigualdade.
Vamos quebrar isso em dois casos:
1)M,N e P coincidem.E ai e so usar Erdös-Mordell em XYZ.
2)MNP e um triangulo.Depois eu continuo.  
ATEEEEEEE!!!!!!loft1Peterdirichlet.
-- Mensagem original --

>PRIMEIRA PARTE:Para aplicar Erdös-Mordell neste problema,e bom que ponhamos
>tudo para dentro!Traçando os paralelogramos DEFM,NFAB,PBCD,e o triangulo
>XYZ tal que ZYe perpendicular a BP,XZ a DM e XY a FN,da para reescrever
>a desigualdade.Prove que
>XM+YN+ZP>=BN+BP+DP+DM+FM+FN.
>Essa e a proxima missao.ATEEEEEEEEEEE!!!!!Ploft!Peterdirichlet
>
>-- Mensagem original --
>
>>Esse problema foi considerado "O Imortal"(o menos respondido de toda a
>historia
>>da IMO).apenas 2 romenos e 4 armenios resolveram-no.TODA A EQUIPE CHINESA
>>ZEROU ESSE.IMO 1996
>>
>> 
>> 
>>Problem 5
>>
>>Let ABCDEF be a convex hexagon such that AB is parallel to DE, BC is parallel
>>to EF, and CD is parallel to FA. Let RA, RC, RE denote the circumradii
>of
>>triangles FAB, BCD, DEF respectively, and let p denote the perimeter of
>>the hexagon. Prove that: 
>>
>>        RA + RC + RE >= p/2. 
>>
>> 
>>
>>Solution
>>
>>
>>The starting point is the formula for the circumradius R of a triangle
>ABC:
>>2R = a/sin A = b/sin B = c/sin C. [Proof: the side a subtends an angle
>2A
>>at the center, so a = 2R sin A.] This gives that 2RA = BF/sin A, 2RC =
>BD/sin
>>C, 2RE = FD/sin E. It is clearly not true in general that BF/sin A > BA
>>+ AF, although it is true if angle FAB >= 120, so we need some argument
>>that involves the hexagon as a whole. 
>>
>>Extend sides BC and FE and take lines perpendicular to them through A
and
>>D, thus forming a rectangle. Then BF is greater than or equal to the side
>>through A and the side through D. We may find the length of the side through
>>A by taking the projections of BA and AF giving AB sin B + AF sin F. Similarly
>>the side through D is CD sin C + DE sin E. Hence: 
>>
>>    2BF >= AB sin B + AF sin F + CD sin C + DE sin E.   Similarly: 
>>
>>    2BD >= BC sin B + CD sin D + AF sin A + EF sin E, and 
>>
>>    2FD >= AB sin A + BC sin C + DE sin D + EF sin F. 
>>
>>Hence 2BF/sin A + 2BD/sin C + 2FD/sin E >= AB(sin A/sin E + sin B/sin
A)
>>+ BC(sin B/sin C + sin C/sin E) + CD(sin C/sin A + sin D/sin C) + DE(sin
>>E/sin A + sin D/sin E) + EF(sin E/sin C + sin F/sin E) + AF(sin F/sin
A
>>+ sin A/sin C). 
>>
>>We now use the fact that opposite sides are parallel, which implies that
>>opposite angles are equal: A = E, B = E, C = F. Each of the factors multiplying
>>the sides in the last expression now has the form x + 1/x which has minimum
>>value 2 when x = 1. Hence 2(BF/sin A + BD/sin C + FD/sin E) >= 2p and
the
>>result is proved. 
>>
>>Essa soluçao e a oficial.A mais bonita e a de Ciprian Manolescu,o unico
>> Perfect Score da prova.Ele usou a famosa Desigualdade de Erdös-Mordell.Depois
>>eu envio a resposta dele.ATEEEEEEEEE.Peterdirichlet  
>>

TRANSIRE SVVM PECTVS MVNDOQUE POTIRE
CONGREGATI EX TOTO ORBE MATHEMATICI OB SCRIPTA INSIGNIA TRIBVERE
Medalha Fields(John Charles Fields)


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